When an RF or microwave signal of excessively high power must be attenuated, one option is to use several cascaded pads. For example, an attenuation of 33 dB could come from a combination of 3 dB and 30 dB. But if the excessive power is really high, attention must be paid to the relative power dissipations of the two pads. In this case, the 3-dB pad would be put in first followed by the 30-dB pad, thus roughly equalizing the power dissipation burden placed on each.
The program given (see the listing) finds what decibel values would be used for equal pad power dissipations in cascades of two to six pads, given the pad cascade's total attenuation. Three hypothetical examples are given to help illustrate. For each total attenuation, those pads which sum to that total are shown along with the fraction of the total input power each pad would absorb. In one case, a total of 10 dB using four pads could be achieved with 1.11, 1.49, 2.28, and 5.12 dB, with each dissipating 0.225 times the total input power. Where odd values of attenuation are called for, the values can serve as a selection guide. In the aforementioned example, 1 dB, 1.5dB, 2.5 dB and 5 dB could be used to achieve a reasonable power dissipation balance.
Next, the user is asked how many decibels of attenuation are needed and with how many pads that attenuation requires to be obtained. Upon entry, the results are presented similarly to the examples (see the figure).
Let Pd be the power dissipated in each attenuator pad (A1 through An), where there are n such pads in cascade. Total attenuation is A dB. Therefore, with each attenuator dissipating Pd:
Pd =(Pin - Pout)/n
The power level at the output of each attenuator Aj is called Pj. At the output of the jth attenuator is:
Pj =P - j(Pd/n)
where j=1 to n.
At the immediately preceding (j - 1)th attenuator is:
Pj - 1 =Pin - (j - 1)(Pd/n)
where j=2 to n.
Therefore, the attenuation of the jth pad, called Aj, is given by:
Aj =10log10\[(Pj - 1)/Pj\].